README_EN.md

290. Word Pattern

中文文档

Description

Given a pattern and a string s, find if s follows the same pattern.

Here follow means a full match, such that there is a bijection between a letter in pattern and a non-empty word in s.

 

Example 1:

Input: pattern = "abba", s = "dog cat cat dog"
Output: true

Example 2:

Input: pattern = "abba", s = "dog cat cat fish"
Output: false

Example 3:

Input: pattern = "aaaa", s = "dog cat cat dog"
Output: false

Example 4:

Input: pattern = "abba", s = "dog dog dog dog"
Output: false

 

Constraints:

• 1 <= pattern.length <= 300
• pattern contains only lower-case English letters.
• 1 <= s.length <= 3000
• s contains only lower-case English letters and spaces ' '.
• s does not contain any leading or trailing spaces.
• All the words in s are separated by a single space.

Solutions

Python3

class Solution:
    def wordPattern(self, pattern: str, s: str) -> bool:
        ch2str, str2ch = {}, {}
        ss = s.split(' ')
        n = len(pattern)
        if n != len(ss):
            return False
        for i in range(n):
            if ch2str.get(pattern[i]) is not None and ch2str.get(pattern[i]) != ss[i]:
                return False
            if str2ch.get(ss[i]) is not None and str2ch.get(ss[i]) != pattern[i]:
                return False
            ch2str[pattern[i]] = ss[i]
            str2ch[ss[i]] = pattern[i]
        return True

Java

class Solution {
    public boolean wordPattern(String pattern, String s) {
        Map<Character, String> ch2str = new HashMap<>();
        Map<String, Character> str2ch = new HashMap<>();
        String[] ss = s.split(" ");
        int n = pattern.length();
        if (n != ss.length) {
            return false;
        }
        for (int i = 0; i < n; ++i) {
            char ch = pattern.charAt(i);
            if (ch2str.containsKey(ch) && !ch2str.get(ch).equals(ss[i])) {
                return false;
            }
            if (str2ch.containsKey(ss[i]) && !str2ch.get(ss[i]).equals(ch)) {
                return false;
            }
            ch2str.put(ch, ss[i]);
            str2ch.put(ss[i], ch);
        }
        return true;
    }
}

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